Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<map>
#define ll long long
#define inf 20021225
#define N 250010
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
const ll mdn = 1000000000000122571ll, bs = 2002122571ll, inv = 938258839498508531ll;
ll mul(ll a,ll b){return (__int128)a*b%mdn;}
ll add(ll a,ll b){return (__int128)(a+b)%mdn;}
map<ll,int> f; char ch[N]; ll ans,val[N],pos[N];
int main()
{
	int n=read(); scanf("%s",ch+1); pos[0]=1;
	for(int i=1;i<=n;i++)
	{
		pos[i]=pos[i-1],val[i]=val[i-1];
		if(ch[i]=='+')	val[i]=add(val[i],pos[i]);
		if(ch[i]=='-')	val[i]=add(val[i],mdn-pos[i]);
		if(ch[i]=='>')	pos[i]=mul(pos[i],bs);
		if(ch[i]=='<')	pos[i]=mul(pos[i],inv);
	}
	for(int i=n;i;i--)
	{
		f[val[i]]++;
		ans+=f[add(mul(val[n],pos[i-1]),val[i-1])];
	}
	printf("%lld\n",ans);
	return 0;
}